Little boy Gerald studies at school which is quite far from his house. That's why he has to go there by bus every day. The way from home to school is represented by a segment of a straight line; the segment contains exactly n + 1 bus stops. All of them are numbered with integers from 0 to n in the order in which they follow from Gerald's home. The bus stop by Gerald's home has number 0 and the bus stop by the school has number n.
There are m buses running between the house and the school: the i-th bus goes from stop si to ti (si < ti), visiting all the intermediate stops in the order in which they follow on the segment. Besides, Gerald's no idiot and he wouldn't get off the bus until it is still possible to ride on it closer to the school (obviously, getting off would be completely pointless). In other words, Gerald can get on the i-th bus on any stop numbered from si to ti - 1 inclusive, but he can get off the i-th bus only on the bus stop ti.
Gerald can't walk between the bus stops and he also can't move in the direction from the school to the house.
Gerald wants to know how many ways he has to get from home to school. Tell him this number. Two ways are considered different if Gerald crosses some segment between the stops on different buses. As the number of ways can be too much, find the remainder of a division of this number by 1000000007 (109 + 7).
The first line contains two space-separated integers: n and m (1 ≤ n ≤ 109, 0 ≤ m ≤ 105). Then follow m lines each containing two integers si, ti. They are the numbers of starting stops and end stops of the buses (0 ≤ si < ti ≤ n).
Print the only number — the number of ways to get to the school modulo 1000000007 (109 + 7).
2 2 0 1 1 2
1
3 2 0 1 1 2
0
5 5 0 1 0 2 0 3 0 4 0 5
16
The first test has the only variant to get to school: first on bus number one to the bus stop number one; then on bus number two to the bus stop number two.
In the second test no bus goes to the third bus stop, where the school is positioned. Thus, the correct answer is 0.
In the third test Gerald can either get or not on any of the first four buses to get closer to the school. Thus, the correct answer is 24 = 16.
题意：有m条公交路线，问你有多少中方案从0到n，每条公交路线的描述为s，t：s为起点，t为终点，可以在除终点外的任意站上车即[s,t-1]间的站，但只能在终点下车。
分析：树状数组+DP，f[t]表示到达t站的方案数，按t对公交路线排序，对于当前的公交车，假设起点站和终点站分别为s,t，那么对于区间[s,t-1]站内上车的都可以到达t，那么查询[s,t-1]之间有的所有方案数的和可以用树状数组求得并维护。由于n>>m所以离散化。
树状数组：
#include<bits/stdc++.h> using namespace std; #define ll long long const int maxn=1e6+5; const ll mod=1e9+7; ll p[maxn],bit[maxn],tol; struct node1 { ll l,r; }c[maxn]; bool cmp(node1 a,node1 b) { if(a.r!=b.r)return a.r<b.r; return a.l<b.l; } ll sum(ll i) { ll s=0; while(i>0) { s=(s+bit[i])%mod; i-=i&-i; } return s%mod; } void add(ll i,ll x) { while(i<=tol) { bit[i]=(bit[i]+x)%mod; i+=i&-i; } } int main() { ll n,m;scanf("%lld%lld",&n,&m); tol=1; for(int i=0;i<m;i++) { scanf("%lld%lld",&c[i].l,&c[i].r); p[tol++]=c[i].l; p[tol++]=c[i].r; } sort(p+1,p+tol+1); sort(c,c+m,cmp); ll s=0; for(int i=0;i<m;i++) { int l=lower_bound(p+1,p+tol+1,c[i].l)-p; int r=lower_bound(p+1,p+tol+1,c[i].r)-p; ll ans=0; if(c[i].l==0)ans++; ans+=sum(r-1)-sum(l-1); ans=(ans+mod)%mod; add(r,ans); if(c[i].r==n)s=sum(r)-sum(r-1); } printf("%lld ",(s+mod)%mod); return 0; }
线段树：
#include<cstdio> #include<string.h> #include<algorithm> using namespace std; #define ll long long const int maxn=1e6+5; const ll mod=1e9+7; ll n,m,p[maxn*2]; struct node1 { ll l,r; }c[maxn]; bool cmp(node1 a,node1 b) { if(a.r!=b.r)return a.r<b.r; return a.l<b.l; } struct node { ll left,right,mid; ll x; }tree[maxn*4]; void build(ll l,ll r,int rt) { tree[rt].left=l; tree[rt].right=r; tree[rt].mid=(l+r)>>1; if(l==r)return; build(l,tree[rt].mid,rt<<1); build(tree[rt].mid+1,r,rt<<1|1); } ll query(ll l,ll r,int rt) { if(l<=tree[rt].left&&r>=tree[rt].right) return tree[rt].x%mod; ll ans=0; if(l<=tree[rt].mid) ans+=query(l,r,rt<<1); ans%=mod; if(r>tree[rt].mid) ans+=query(l,r,rt<<1|1); return ans%mod; } void add(ll L,ll C,int rt) { if(tree[rt].left==tree[rt].right) { tree[rt].x=(tree[rt].x+C)%mod; return; } if(L<=tree[rt].mid) add(L,C,rt<<1); else add(L,C,rt<<1|1); tree[rt].x=(tree[rt<<1].x+tree[rt<<1|1].x)%mod; } int main() { scanf("%lld%lld",&n,&m); int tol=1; for(int i=0;i<m;i++) { scanf("%lld%lld",&c[i].l,&c[i].r); p[tol++]=c[i].l; p[tol++]=c[i].r; } sort(p+1,p+tol+1); sort(c,c+m,cmp); build(1,tol,1); ll sum=0; for(int i=0;i<m;i++) { ll ans=0; ll l=lower_bound(p+1,p+tol+1,c[i].l)-p; ll r=lower_bound(p+1,p+tol+1,c[i].r)-p; if(c[i].l==0)ans++; if(r>=l)ans+=query(l,r-1,1); add(r,ans,1); if(c[i].r==n)sum=query(r,r,1); } printf("%lld ",sum); return 0; }